# Why/how normal shear stress is zero in thin-walled beams?

Please explain how the normal shear stress (shear stress normal to the cuve, C, defining the cross-section of a thin-walled beam) is zero. I have attached a screenshot from a book which says that the outer surfaces of the beam are stress free. This cannot be true except in very special cases. Right? Now consider the following image. Here axis $$1$$ is the axial direction of the beam, and axes $$2$$ and $$3$$ define the cross-section of the beam. For the normal shear stress to be zero in the member $$A$$ of the cross-section, the shear stress $$\sigma_{21}$$ should be zero on the lateral face $$1$$ which is possible only if there is no axial force on the lateral face $$1$$. Am I right? • I suspect the assumption is that the member is infinitely thin, and being infinitely thin, would have infinite curvature with any finite shear stress. In a more real sense, the thin walls are supported in such a way that any shear stress across the thin section is (we hope) negligible (because again, if they weren't, they'd bend disproportionately to the rest of the beam). Mar 31 '19 at 1:50

In the diagram in the book, the stress component $$\tau_n$$ will be zero at the surface if there is no pressure load on the surface at that position along the beam. Otherwise, it will not be zero.
The book says it is assumed that "$$\tau_n$$ is zero through the wall thickness." That is an assumption, and it is only an approximation. In fact, for a uniform beam there is a parabolic variation of $$\tau_n$$ through the thickness. This can be ignored for a slender beam (which is thin relative to its length) because it is small compared with the other stress components, but it is not zero. For a thick beam, this shear stress component is included in the beam formulation.
• Equation 1.5 is about reciprocity of shear stress, i.e., $\sigma_{ij}= \sigma_{ji}$. Since we are dealing with a Euler-Bernoulli beam, the only non-vanishing stress components are $\sigma_1$, $\sigma_{12}$, and $\sigma_{13}$, with 1 being the axial direction. Does it makes sense now? If yes, please explain. Mar 30 '19 at 10:54